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In how many different ways can we choose $12$ cans of soup if there are $5$ different varieties available, if

  1. the order of the chosen cans does not count?
  2. the order of the chosen cans does count?

I have tried as follows:

  1. Since the order does not matter I used a combination to obtain: $792$
  2. by permutation: $95040$

Please confirm whether I'm on the right track or not. Thank you.

N. F. Taussig
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user402194
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  • My guess is that in the first question, you calculated comb(12,5) = 792. I don't think this is correct. Read this: https://math.stackexchange.com/a/910820/432405 – Matti P. Sep 17 '18 at 12:20

1 Answers1

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For 1), what you have done is choose 5 cans if there are 12 variations available (and no repetition allowed). The way i read the question, that's not what you want. The way i read it, there are five varieties, unlimited supply of each, and you want to pick 12 cans with or without ordering, but with possible repeats.

With ordering is the easiest: For each of the $12$ cas you choose, you have $5$ alternatives, so the total number of different choices is $5^{12}$.

Without ordering is a bit more tricky, and is most commonly solved using a technique known as stars and bars. The idea is that you have a shelf with designated spost for each of the five types of cans, and once you've chosen your twelve cans, you put them in their designated spots on the shelf and study them. You can describe one choice of twelve cans uniquely by just saying how many are placed in each spot on the shelf. Here is a very simplified picture that shelf, with one way to choose cans: $$ {}*{}*{}*{}\mid{}*{}*{}*{}\mid{}*{}*{}*{}*{}\mid{}*{}\mid{}*{} $$ This symbolizes choosing three cans of type 1, three of type 2, four of type 3, and one of each of types 4 and 5.

Since each possible row of four dividers ("bars") and twelve cans ("stars") corresponds to one way of choosing 12 cans (and vice versa), we have that the number of way to choose cans is equal to the number of possible orderings of those symbols.

There are $16$ symbols, and once you've chosen $12$ of them to be stars, the remaining $4$ must be bars. Thus this can be done in $\binom{16}{12}$ ways.

Arthur
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